Simplify the following expression and state the condition under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{3n(2n - 3)}{3n} \times \dfrac{2}{16n - 24} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ 3n(2n - 3) \times 2 } { 3n \times (16n - 24) } $ $ r = \dfrac {2 \times 3n(2n - 3)} {3n \times 8(2n - 3)} $ $ r = \dfrac{6n(2n - 3)}{24n(2n - 3)} $ We can cancel the $2n - 3$ so long as $2n - 3 \neq 0$ Therefore $n \neq \dfrac{3}{2}$ $r = \dfrac{6n \cancel{(2n - 3})}{24n \cancel{(2n - 3)}} = \dfrac{6n}{24n} = \dfrac{1}{4} $